Margulis Lemma

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18 VITALI KAPOVITCH AND BURKHARD WILKING Using the definition of dt r (s) and the volume estimate (9) this gives ∫ − dt r (s)(p, q) dµ(p) dµ(q) ≤ C 2 (n) · r · ε + 2rC 1 ε =: C 3 rε. B r(c(t)) 2 This completes the induction step with C(n) = C 3 and ε 0 = 1 2C 3 . In order to remove the restriction ε ≤ ε 0 one can just increase C 3 by the factor 4, as indicated at the beginning. □ Proof of Proposition 3.6. Put g t,i (x) = ‖∇·Xi t ‖(x). First notice that by Hölder ∫ 1 0 Mx(g t,i (x))(c i (t)) dt ≤ ∫ 1 0 u t,i (c i (t)) dt → 0. Let λ i → ∞ and put r i = R λ i , where R > 1 is arbitrary. By Lemma 3.7 there is S i ⊂ B ri (c i (0)) with • vol(S i ) ≥ (1 − δ i ) vol(B ri (c i (0))) with δ i → 0 and • φ t (S i ) ⊂ B 2ri (c(t)) for all t. In the following we assume that i is so large that δ i ≤ 1/2, and r i ≤ 1/100. As in the proof of Lemma 3.7 this easily implies that there is a universal constant C = C(n) with and Using that φ it|Bri (c(0)) ′ vol(B 2ri (c i(t))) vol(B ri (c i(0))) ≤ C 2 for all t and all i. is measure preserving, we deduce ∫ 1 − Mx(g t,i )(φ it (p)) dt dµ(p) ≤ C ∫S i 0 ≤ by (5) C ∫ 1 0 ∫ 1 0 ∫ − B 2ri (c i(t)) Mx(g t,i )(q) dt dµ(q) Mx(Mx(g t,i ))(c i (t)) dt ∫ 1 ≤ C · C 2 Mx(gt,i) α 1/α (c i (t)) dt 0 ∫ 1 = C · C 2 u t,i (c i (t)) dt → 0. Thus we can find ˜δ i → 0 and a subset B ri (c i (0)) ′′ ⊂ S i with vol(B ri (c i (0))) − vol(B ri (c i (0)) ′′ ) ≤ ˜δ i min vol(B r q∈B ri (c i/R(q)) i(0)) ∫ 1 0 Mx(g t,i )(φ it (q)) ≤ ˜δ i for all q ∈ B ri (c(0)) ′′ . Recall that r i = R λ i with an arbitrary R. By a diagonal sequence argument it is easy to deduce that after replacing ˜δ i by a another sequence converging slowly to 0 we can keep the above estimates for r i = Ri λ i with R i → ∞ sufficiently slowly. Combining this with Lemma 3.7 shows that f i = φ i1 : (λ i M i , c(0)) → (λ i M i , c(1)) has the zooming in property. Let ˜X i t be a lift of Xt i to the universal covering ˜M i of M i . Consider the integral curve ˜c i : [0, 1] → ˜M i of ˜Xt i with ˜c(0) = ˜p i . Clearly ˜c i is a lift of c i and by the 0
STRUCTURE OF FUNDAMENTAL GROUPS 19 Covering Lemma (1.6) we have ∫ 1 0 Mx(‖∇· ˜Xt i ‖ α ) 1/α (˜c i (t)) dt ≤ C ∫ 1 0 Mx(‖∇·X t i ‖ α ) 1/α (c i (t)) dt with some universal constant C. Thus ˜φ i1 : (λ i ˜Mi , ˜p i ) → (λ i ˜Mi , φ i1 (˜p i )) has the zooming in property as well. Any other lift ˜f i of f i is obtained by composing ˜φ i1 with a deck transformation and thus the result carries over to any lift of f i . □ Proposition 3.8 (Second main example). Let (M i , g i ) be a sequence of manifolds with Ric > −1/i on B i (p i ) and B i (p i ) compact. Suppose that (M i , p i ) converges to (R k × Y, p ∞ ). Then for each v ∈ R k there is a sequence of diffeomorphisms f i : [M i , p i ] → [M i , p i ] with the zooming in property which converges in the weakly measured sense to an isometry f ∞ of R k × Y that acts trivially on Y and by w ↦→ w + v on R k . Moreover, f i is isotopic to the identity and there is a lift ˜f i : [ ˜M i , ˜p i ] → [ ˜M i , ˜p i ] of f i to the universal cover which has the zooming in property as well. Proof. Using the splitting R k = Rv ⊕(v) ⊥ and replacing Y by Y ×(v) ⊥ we see that it suffices to prove the statement for k = 1. By the work of Cheeger and Colding [CC96] we can find sequences ρ i → ∞ and ε i → 0 and harmonic functions b i : B 4ρi (p i ) → R such that |∇b i | ≤ L(n) for all i and ∫ (∣ − ∣|∇bi | − 1 ∣ + ‖Hessbi ‖ ) 2 ≤ ε i for any R ∈ [1/4, 4ρ i ]. B 4R (p i) Let X i be a vector field with compact support with X i = ∇b i on B 3ρi (p i ), and let φ it denote the flow of X i . Clearly for any t we can find r i → ∞ such that φ it|Bri (p i) is measure preserving. Put ψ i := ∣ ∣|∇b i | − 1 ∣ + ‖Hess(b i )‖. We deduce from Lemma 1.4 that ∫ − Mx(ψ i ) 2 ≤ C(n, R)ε i and Cauchy Schwarz gives ∫ − B 2R (p i) B R (p i) Mx(ψ i ) ≤ √ C(n, R)ε i . Suppose now that t 0 ≤ R 4L(n) . Then φ t(q) ∈ B 3R (p i ) for all q ∈ B 4 R/2 (p i ) and all t ∈ [−t 0 , t 0 ]. Combining that φ t is measure preserving and vol(B R (p i )) ≤ C 3 vol(B R/2 (p i )), we get that ∫ ∫ t0 ∫ − Mx(ψ i )(φ t (p)) ≤ C 3 t 0 − Mx(ψ) B R/2 (p i) 0 It is now easy to find R i → ∞ and δ i → 0 with B R (p i) ≤ C 4 (t 0 , R, n) √ ε i . ∫ ∫ t0 − Mx(ψ i )(φ t (p)) ≤ δ i for all R ∈ [1, R i ]. B R (p i) 0
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Margulis Lemma

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